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Question Number 24010    Answers: 0   Comments: 0

Prove that Σ_(r=1) ^(2n−1) (−1)^(r−1) ∙(r/(^(2n) C_r )) = (n/(n + 1)) .

$$\mathrm{Prove}\:\mathrm{that}\:\underset{{r}=\mathrm{1}} {\overset{\mathrm{2}{n}−\mathrm{1}} {\sum}}\left(−\mathrm{1}\right)^{{r}−\mathrm{1}} \centerdot\frac{{r}}{\:^{\mathrm{2}{n}} {C}_{{r}} }\:=\:\frac{{n}}{{n}\:+\:\mathrm{1}}\:. \\ $$

Question Number 23984    Answers: 2   Comments: 0

Which of the following relation is/are correct? (1) ΔG = ΔH − TΔS (2) ΔG = ΔH + T[((δ(ΔG))/(δT))]_P (3) ΔG = ΔH + TΔS (4) ΔG = ΔH + ΔnRT

$$\mathrm{Which}\:\mathrm{of}\:\mathrm{the}\:\mathrm{following}\:\mathrm{relation}\:\mathrm{is}/\mathrm{are} \\ $$$$\mathrm{correct}? \\ $$$$\left(\mathrm{1}\right)\:\Delta\mathrm{G}\:=\:\Delta\mathrm{H}\:−\:\mathrm{T}\Delta\mathrm{S} \\ $$$$\left(\mathrm{2}\right)\:\Delta\mathrm{G}\:=\:\Delta\mathrm{H}\:+\:\mathrm{T}\left[\frac{\delta\left(\Delta\mathrm{G}\right)}{\delta\mathrm{T}}\right]_{\mathrm{P}} \\ $$$$\left(\mathrm{3}\right)\:\Delta\mathrm{G}\:=\:\Delta\mathrm{H}\:+\:\mathrm{T}\Delta\mathrm{S} \\ $$$$\left(\mathrm{4}\right)\:\Delta\mathrm{G}\:=\:\Delta\mathrm{H}\:+\:\Delta\mathrm{nRT} \\ $$

Question Number 23960    Answers: 1   Comments: 4

Question Number 23957    Answers: 0   Comments: 0

∫((sinx−2cosx)/(1+2sin2x))dx solves

$$\int\frac{\boldsymbol{\mathrm{sinx}}−\mathrm{2}\boldsymbol{\mathrm{cosx}}}{\mathrm{1}+\mathrm{2}\boldsymbol{\mathrm{sin}}\mathrm{2}\boldsymbol{\mathrm{x}}}\boldsymbol{\mathrm{dx}} \\ $$$$\boldsymbol{\mathrm{solves}} \\ $$

Question Number 23968    Answers: 0   Comments: 3

Prove that n and 2n−1 are coprime.

$${Prove}\:{that}\:{n}\:{and}\:\mathrm{2}{n}−\mathrm{1}\:{are}\:{coprime}. \\ $$

Question Number 23972    Answers: 0   Comments: 0

∫_0 ^(90) ((sin^2 x)/(sin^4 x+cos^4 x))dx

$$\int_{\mathrm{0}} ^{\mathrm{90}} \frac{\mathrm{sin}\:^{\mathrm{2}} {x}}{\mathrm{sin}\:^{\mathrm{4}} {x}+\mathrm{cos}\:^{\mathrm{4}} {x}}{dx} \\ $$

Question Number 23953    Answers: 0   Comments: 4

find the nth term of the sequence 4, 9, 16, 45, 76 please help

$${find}\:{the}\:{nth}\:{term}\:{of}\:{the}\:{sequence} \\ $$$$\mathrm{4},\:\mathrm{9},\:\mathrm{16},\:\mathrm{45},\:\mathrm{76} \\ $$$$ \\ $$$${please}\:{help} \\ $$

Question Number 23940    Answers: 0   Comments: 2

When the gas is ideal and process is isothermal, then (1) P_1 V_1 = P_2 V_2 (2) ΔU = 0 (3) ΔW = 0 (4) ΔH_1 = ΔH_2

$$\mathrm{When}\:\mathrm{the}\:\mathrm{gas}\:\mathrm{is}\:\mathrm{ideal}\:\mathrm{and}\:\mathrm{process}\:\mathrm{is} \\ $$$$\mathrm{isothermal},\:\mathrm{then} \\ $$$$\left(\mathrm{1}\right)\:\mathrm{P}_{\mathrm{1}} \mathrm{V}_{\mathrm{1}} \:=\:\mathrm{P}_{\mathrm{2}} \mathrm{V}_{\mathrm{2}} \\ $$$$\left(\mathrm{2}\right)\:\Delta\mathrm{U}\:=\:\mathrm{0} \\ $$$$\left(\mathrm{3}\right)\:\Delta\mathrm{W}\:=\:\mathrm{0} \\ $$$$\left(\mathrm{4}\right)\:\Delta\mathrm{H}_{\mathrm{1}} \:=\:\Delta\mathrm{H}_{\mathrm{2}} \\ $$

Question Number 23939    Answers: 0   Comments: 0

In a school of 500 students,150 played hockey 250 played football,120 played table tennis and 100 played one of the three games.If 50 played both hockey and football,how many played table tennis only?pls show workings with diagram

$${In}\:{a}\:{school}\:{of}\:\mathrm{500}\:{students},\mathrm{150}\:{played}\:{hockey} \\ $$$$\mathrm{250}\:{played}\:{football},\mathrm{120}\:{played}\:{table}\:{tennis}\:{and} \\ $$$$\mathrm{100}\:{played}\:{one}\:{of}\:{the}\:{three}\:{games}.{If}\:\mathrm{50}\: \\ $$$${played}\:{both}\:{hockey}\:{and}\:{football},{how}\:{many} \\ $$$${played}\:{table}\:{tennis}\:{only}?{pls}\:{show}\:{workings}\:{with} \\ $$$${diagram} \\ $$$$ \\ $$

Question Number 23937    Answers: 1   Comments: 1

Question Number 23928    Answers: 0   Comments: 4

Find^n C_1 − (1/2)^n C_2 + (1/3)^n C_3 − .... + (−1)^(n−1) (1/n) ∙^n C_n .

$$\mathrm{Find}\:^{{n}} {C}_{\mathrm{1}} \:−\:\frac{\mathrm{1}}{\mathrm{2}}\:^{{n}} {C}_{\mathrm{2}} \:+\:\frac{\mathrm{1}}{\mathrm{3}}\:^{{n}} {C}_{\mathrm{3}} \:−\:....\:+ \\ $$$$\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \frac{\mathrm{1}}{{n}}\:\centerdot\:^{{n}} {C}_{{n}} . \\ $$

Question Number 23932    Answers: 1   Comments: 0

A cord is wound around the circumference of a bicycle wheel (without tyre) of diameter 1 m. A mass of 2 kg is tied at the end of the cord and it is allowed to fall from rest. The weight falls 2 m in 4 s. The axle of the wheel is horizontal and the wheel rotates with its plane vertical. If g = 10 ms^(−2) , what is the angular acceleration of the wheel?

$$\mathrm{A}\:\mathrm{cord}\:\mathrm{is}\:\mathrm{wound}\:\mathrm{around}\:\mathrm{the}\:\mathrm{circumference} \\ $$$$\mathrm{of}\:\mathrm{a}\:\mathrm{bicycle}\:\mathrm{wheel}\:\left(\mathrm{without}\:\mathrm{tyre}\right)\:\mathrm{of} \\ $$$$\mathrm{diameter}\:\mathrm{1}\:\mathrm{m}.\:\mathrm{A}\:\mathrm{mass}\:\mathrm{of}\:\mathrm{2}\:\mathrm{kg}\:\mathrm{is}\:\mathrm{tied}\:\mathrm{at} \\ $$$$\mathrm{the}\:\mathrm{end}\:\mathrm{of}\:\mathrm{the}\:\mathrm{cord}\:\mathrm{and}\:\mathrm{it}\:\mathrm{is}\:\mathrm{allowed}\:\mathrm{to} \\ $$$$\mathrm{fall}\:\mathrm{from}\:\mathrm{rest}.\:\mathrm{The}\:\mathrm{weight}\:\mathrm{falls}\:\mathrm{2}\:\mathrm{m}\:\mathrm{in}\:\mathrm{4} \\ $$$$\mathrm{s}.\:\mathrm{The}\:\mathrm{axle}\:\mathrm{of}\:\mathrm{the}\:\mathrm{wheel}\:\mathrm{is}\:\mathrm{horizontal} \\ $$$$\mathrm{and}\:\mathrm{the}\:\mathrm{wheel}\:\mathrm{rotates}\:\mathrm{with}\:\mathrm{its}\:\mathrm{plane} \\ $$$$\mathrm{vertical}.\:\mathrm{If}\:\mathrm{g}\:=\:\mathrm{10}\:\mathrm{ms}^{−\mathrm{2}} ,\:\mathrm{what}\:\mathrm{is}\:\mathrm{the} \\ $$$$\mathrm{angular}\:\mathrm{acceleration}\:\mathrm{of}\:\mathrm{the}\:\mathrm{wheel}? \\ $$

Question Number 23922    Answers: 1   Comments: 0

if f(x)=2^x show that f(x+3)−f(x−1)=((15)/2)f(x)

$$\boldsymbol{\mathrm{if}}\:\boldsymbol{\mathrm{f}}\left(\boldsymbol{\mathrm{x}}\right)=\mathrm{2}^{\boldsymbol{\mathrm{x}}} \\ $$$$\boldsymbol{\mathrm{show}}\:\boldsymbol{\mathrm{that}}\:\boldsymbol{\mathrm{f}}\left(\boldsymbol{\mathrm{x}}+\mathrm{3}\right)−\boldsymbol{\mathrm{f}}\left(\boldsymbol{\mathrm{x}}−\mathrm{1}\right)=\frac{\mathrm{15}}{\mathrm{2}}\boldsymbol{\mathrm{f}}\left(\boldsymbol{\mathrm{x}}\right) \\ $$

Question Number 23920    Answers: 1   Comments: 1

Question Number 23981    Answers: 1   Comments: 0

Amount of heat required to change 1 g ice at 0°C to 1 g steam at 100°C is (1) 616 cal (2) 12 kcal (3) 717 cal (4) none of these.

$$\mathrm{Amount}\:\mathrm{of}\:\mathrm{heat}\:\mathrm{required}\:\mathrm{to}\:\mathrm{change}\:\mathrm{1}\:{g} \\ $$$$\mathrm{ice}\:\mathrm{at}\:\mathrm{0}°\mathrm{C}\:\mathrm{to}\:\mathrm{1}\:{g}\:\mathrm{steam}\:\mathrm{at}\:\mathrm{100}°\mathrm{C}\:\mathrm{is} \\ $$$$\left(\mathrm{1}\right)\:\mathrm{616}\:\mathrm{cal} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{12}\:\mathrm{kcal} \\ $$$$\left(\mathrm{3}\right)\:\mathrm{717}\:\mathrm{cal} \\ $$$$\left(\mathrm{4}\right)\:\mathrm{none}\:\mathrm{of}\:\mathrm{these}. \\ $$

Question Number 23925    Answers: 0   Comments: 0

Question Number 23900    Answers: 0   Comments: 0

Question Number 23897    Answers: 1   Comments: 3

Question Number 23891    Answers: 0   Comments: 3

Question Number 23883    Answers: 0   Comments: 0

plz anyone answer the question 23870 .plz plz plz.....

$${plz}\:{anyone}\:\:\:{answer}\:{the}\:{question}\:\mathrm{23870}\:.{plz}\:{plz}\:{plz}..... \\ $$

Question Number 23871    Answers: 0   Comments: 0

Let ABC be a triangle and B′ be the reflection of B in the line CA and C′ be reflection of C in the line AB. Prove that ΔABC′ ≅ ΔACB′ ≅ ΔABC.

$$\mathrm{Let}\:{ABC}\:\mathrm{be}\:\mathrm{a}\:\mathrm{triangle}\:\mathrm{and}\:{B}'\:\mathrm{be}\:\mathrm{the} \\ $$$$\mathrm{reflection}\:\mathrm{of}\:{B}\:\mathrm{in}\:\mathrm{the}\:\mathrm{line}\:{CA}\:\mathrm{and}\:{C}'\:\mathrm{be} \\ $$$$\mathrm{reflection}\:\mathrm{of}\:{C}\:\mathrm{in}\:\mathrm{the}\:\mathrm{line}\:{AB}.\:\mathrm{Prove} \\ $$$$\mathrm{that}\:\Delta{ABC}'\:\cong\:\Delta{ACB}'\:\cong\:\Delta{ABC}. \\ $$

Question Number 23870    Answers: 0   Comments: 0

Find a vector function F whose graph is the the line of intersection of the plane 4x−2y+z=3 and x+y−3y=1

$${Find}\:{a}\:{vector}\:{function}\:{F}\:\:{whose}\:{graph}\:{is}\:{the} \\ $$$${the}\:{line}\:{of}\:{intersection}\:{of}\:{the}\:{plane}\: \\ $$$$\mathrm{4}{x}−\mathrm{2}{y}+{z}=\mathrm{3}\:{and}\:{x}+{y}−\mathrm{3}{y}=\mathrm{1} \\ $$

Question Number 23884    Answers: 1   Comments: 0

If C_r stands for^n C_r = ((n!)/(r! n − r!)) and Σ_(r=1) ^n r.C_r ^2 = λ for n ≥ 2, then λ is divisible by (1) 3 (n − 1) (2) n + 1 (3) n (2n − 1) (4) n^2 + 1

$$\mathrm{If}\:{C}_{{r}} \:\mathrm{stands}\:\mathrm{for}\:^{{n}} {C}_{{r}} \:=\:\frac{{n}!}{{r}!\:{n}\:−\:{r}!}\:\mathrm{and} \\ $$$$\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}{r}.{C}_{{r}} ^{\mathrm{2}} \:=\:\lambda\:\mathrm{for}\:{n}\:\geqslant\:\mathrm{2},\:\mathrm{then}\:\lambda\:\mathrm{is}\:\mathrm{divisible} \\ $$$$\mathrm{by} \\ $$$$\left(\mathrm{1}\right)\:\mathrm{3}\:\left({n}\:−\:\mathrm{1}\right) \\ $$$$\left(\mathrm{2}\right)\:{n}\:+\:\mathrm{1} \\ $$$$\left(\mathrm{3}\right)\:{n}\:\left(\mathrm{2}{n}\:−\:\mathrm{1}\right) \\ $$$$\left(\mathrm{4}\right)\:{n}^{\mathrm{2}} \:+\:\mathrm{1} \\ $$

Question Number 23865    Answers: 0   Comments: 7

A man of mass m, standing at the bottom of the staircase of height L climbs it and stands at its top. (1) Work done by all forces on man is equal to the rise in potential energy mgL. (2) Work done by all forces on man is zero. (3) Work done by the gravitational force on man is mgL. (4) The reaction force from a step does not do work because the point of application of the force does not move while the force exists.

$$\mathrm{A}\:\mathrm{man}\:\mathrm{of}\:\mathrm{mass}\:{m},\:\mathrm{standing}\:\mathrm{at}\:\mathrm{the} \\ $$$$\mathrm{bottom}\:\mathrm{of}\:\mathrm{the}\:\mathrm{staircase}\:\mathrm{of}\:\mathrm{height}\:{L} \\ $$$$\mathrm{climbs}\:\mathrm{it}\:\mathrm{and}\:\mathrm{stands}\:\mathrm{at}\:\mathrm{its}\:\mathrm{top}. \\ $$$$\left(\mathrm{1}\right)\:\mathrm{Work}\:\mathrm{done}\:\mathrm{by}\:\mathrm{all}\:\mathrm{forces}\:\mathrm{on}\:\mathrm{man}\:\mathrm{is} \\ $$$$\mathrm{equal}\:\mathrm{to}\:\mathrm{the}\:\mathrm{rise}\:\mathrm{in}\:\mathrm{potential}\:\mathrm{energy} \\ $$$${mgL}. \\ $$$$\left(\mathrm{2}\right)\:\mathrm{Work}\:\mathrm{done}\:\mathrm{by}\:{all}\:\mathrm{forces}\:\mathrm{on}\:\mathrm{man}\:\mathrm{is} \\ $$$$\mathrm{zero}. \\ $$$$\left(\mathrm{3}\right)\:\mathrm{Work}\:\mathrm{done}\:\mathrm{by}\:\mathrm{the}\:\mathrm{gravitational} \\ $$$$\mathrm{force}\:\mathrm{on}\:\mathrm{man}\:\mathrm{is}\:{mgL}. \\ $$$$\left(\mathrm{4}\right)\:\mathrm{The}\:\mathrm{reaction}\:\mathrm{force}\:\mathrm{from}\:\mathrm{a}\:\mathrm{step}\:\mathrm{does} \\ $$$$\mathrm{not}\:\mathrm{do}\:\mathrm{work}\:\mathrm{because}\:\mathrm{the}\:\mathrm{point}\:\mathrm{of} \\ $$$$\mathrm{application}\:\mathrm{of}\:\mathrm{the}\:\mathrm{force}\:\mathrm{does}\:\mathrm{not}\:\mathrm{move} \\ $$$$\mathrm{while}\:\mathrm{the}\:\mathrm{force}\:\mathrm{exists}. \\ $$

Question Number 23863    Answers: 0   Comments: 0

From the following data, calculate the enthalpy change for the combustion of cyclopropane at 298 K. The enthalpy of formation of CO_2 (g), H_2 O (l) and propene (g) are −393.5, −285.8 and 20.42 kJ mol^(−1) respectively. The enthalpy of isomerisation of cyclopropane to propene is −33 kJ mol^(−1) .

$$\mathrm{From}\:\mathrm{the}\:\mathrm{following}\:\mathrm{data},\:\mathrm{calculate}\:\mathrm{the} \\ $$$$\mathrm{enthalpy}\:\mathrm{change}\:\mathrm{for}\:\mathrm{the}\:\mathrm{combustion}\:\mathrm{of} \\ $$$$\mathrm{cyclopropane}\:\mathrm{at}\:\mathrm{298}\:\mathrm{K}.\:\mathrm{The}\:\mathrm{enthalpy} \\ $$$$\mathrm{of}\:\mathrm{formation}\:\mathrm{of}\:\mathrm{CO}_{\mathrm{2}} \:\left({g}\right),\:\mathrm{H}_{\mathrm{2}} \mathrm{O}\:\left({l}\right)\:\mathrm{and} \\ $$$$\mathrm{propene}\:\left({g}\right)\:\mathrm{are}\:−\mathrm{393}.\mathrm{5},\:−\mathrm{285}.\mathrm{8}\:\mathrm{and} \\ $$$$\mathrm{20}.\mathrm{42}\:\mathrm{kJ}\:\mathrm{mol}^{−\mathrm{1}} \:\mathrm{respectively}.\:\mathrm{The} \\ $$$$\mathrm{enthalpy}\:\mathrm{of}\:\mathrm{isomerisation}\:\mathrm{of}\:\mathrm{cyclopropane} \\ $$$$\mathrm{to}\:\mathrm{propene}\:\mathrm{is}\:−\mathrm{33}\:\mathrm{kJ}\:\mathrm{mol}^{−\mathrm{1}} . \\ $$

Question Number 23856    Answers: 0   Comments: 4

The value of (C_0 + C_1 )(C_1 + C_2 ).... (C_(n−1) + C_n ) is (1) (((n + 1)^n )/(n!)) ∙ C_1 C_2 .....C_n (2) (((n − 1)^n )/(n!)) ∙ C_1 C_2 .....C_n (3) (((n)^n )/((n + 1)!)) ∙ C_1 C_2 .....C_n (4) (((n)^n )/(n!)) ∙ C_1 C_2 .....C_n

$$\mathrm{The}\:\mathrm{value}\:\mathrm{of}\:\left({C}_{\mathrm{0}} \:+\:{C}_{\mathrm{1}} \right)\left({C}_{\mathrm{1}} \:+\:{C}_{\mathrm{2}} \right).... \\ $$$$\left({C}_{{n}−\mathrm{1}} \:+\:{C}_{{n}} \right)\:\mathrm{is} \\ $$$$\left(\mathrm{1}\right)\:\frac{\left({n}\:+\:\mathrm{1}\right)^{{n}} }{{n}!}\:\centerdot\:{C}_{\mathrm{1}} {C}_{\mathrm{2}} .....{C}_{{n}} \\ $$$$\left(\mathrm{2}\right)\:\frac{\left({n}\:−\:\mathrm{1}\right)^{{n}} }{{n}!}\:\centerdot\:{C}_{\mathrm{1}} {C}_{\mathrm{2}} .....{C}_{{n}} \\ $$$$\left(\mathrm{3}\right)\:\frac{\left({n}\right)^{{n}} }{\left({n}\:+\:\mathrm{1}\right)!}\:\centerdot\:{C}_{\mathrm{1}} {C}_{\mathrm{2}} .....{C}_{{n}} \\ $$$$\left(\mathrm{4}\right)\:\frac{\left({n}\right)^{{n}} }{{n}!}\:\centerdot\:{C}_{\mathrm{1}} {C}_{\mathrm{2}} .....{C}_{{n}} \\ $$

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